1. No category

Physics 5300, Theoretical Mechanics Spring 2015
Assignment 6 solutions
The problems numbers below are from Classical Mechanics, John R. Taylor, University
Science Books (2005).
Problem 1
Taylor 13.1
Solution:
1
L = mx˙ 2
2
∂L
= mx˙
p=
∂ x˙
1
1
p2
H = px˙ − L = mx˙ 2 − mx˙ 2 = mx˙ 2 =
2
2
2m
∂H
p
x˙ =
=
∂p
m
∂H
=0
∂x
p0
p0
x˙ = , x = t + x0
m
m
p˙ = −
p = mx˙ = p0 ,
Problem 2
Solution:
(1)
(2)
(3)
(4)
(5)
(6)
Taylor 13.2
Using x oriented downwards, we have
1
L = mx˙ 2 + mgx
2
(7)
p = mx˙
(8)
p2
− mgx
2m
p
x˙ =
m
p˙ = mg
H=
1
(9)
(10)
(11)
Problem 3
Taylor 13.5
Solution:
1
1
1
T = m[z˙ 2 + R2 φ˙ 2 ] = m[c2 φ˙ 2 + R2 φ˙ 2 ] = m[c2 + R2 ]φ˙ 2
2
2
2
(12)
V = mgz = mgcφ
(13)
1
L = T − V = m[c2 + R2 ]φ˙ 2 − mgcφ
2
p = m[c2 + R2 ]φ˙
(16)
∂H
p
φ˙ =
=
2
∂p
m[c + R2 ]
(17)
m[c2 + R2 ]φ¨ = −mgc,
∂H
= −mgc
∂φ
φ¨ = −
gc
,
2
[c + R2 ]
(18)
z¨ = −
gc2
[c2 + R2 ]
(19)
Taylor 13.9
Solution:
1
L = m[x˙ 2 + y˙ 2 ] − mgy
2
px = mx,
˙ py = my˙
H=
p2y
p2x
+
+ mgy
2m 2m
px
x˙ =
m
py
y˙ =
m
p˙x = 0
p˙y = −mg
Problem 5
(15)
p2
1
+ mgcφ
H = pφ˙ − L = m[c2 + R2 ]φ˙ 2 + mgcφ =
2
2m[c2 + R2 ]
p˙ = −
Problem 4
(14)
Taylor 13.26
2
(20)
(21)
(22)
(23)
(24)
(25)
(26)
Solution:
We have
dV
kx4
, V =
dx
4
where we can set an arbitrary additive constant in V to zero.
F = −kx3 = −
(27)
1
kx4
L = mx˙ 2 −
2
4
(28)
p = mx˙
(29)
H=
p2
kx4
+
2m
4
x˙ = p
(30)
(31)
p˙ = −kx3
(32)
p
x
Figure 1: Phase space orbits.
Problem 6
Solution:
Taylor 13.28
1
V = − kx2
2
1
1
L = mx˙ 2 + kx2
2
2
p = mx˙
F = kx,
3
(33)
(34)
(35)
H=
For
p2
1
− kx2
2m 2
p
x˙ =
m
p˙ = kx
(36)
(37)
(38)
p2
1
− kx2 = E < 0
2m 2
we have
x2 =
(39)
2 p2
[
+ |E|]
k 2m
(40)
Since p ≥ 0, we have
r
2|E|
k
The orbits come from |x| = ∞ and return to |x| = ∞.
For
p2
1
− kx2 = E > 0
2m 2
we have
1
p2 = 2m[ kx2 + |E|]
2
Thus
√
|p| ≥ 2mE
|x| ≥
(41)
(42)
(43)
(44)
Thus a particle travelling to the right keeps travelling to the right, while one travelling to
the left keeps travelling to the left.
U(x)
p
x
x
(a)
(b)
Figure 2: (a) The potential (b) Phase space orbits.
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